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iogann1982 [59]
11 months ago
12

A block of mass 200g is oscillating on the end of a horizontal spring of spring constant 100 N/m and natural length 12 cm. When

the spring is stretched to a length of 14 cm, what is the acceleration of the block? [F=ma=kx]
Physics
1 answer:
malfutka [58]11 months ago
8 0

In order to determine the acceleration of the block, use the following formula:

F=ma

Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

F=kx

Then, you have:

ma=kx

by solving for a, you obtain:

a=\frac{kx}{m}

In this case, you have:

k: spring constant = 100N/m

m: mass of the block = 200g = 0.2kg

x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m

Replace the previous values of the parameters into the expression for a:

a=\frac{(\frac{100N}{m})(0.02m)}{0.2\operatorname{kg}}=10\frac{m}{s^2}

Hence, the acceleration of the block is 10 m/s^2

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Light is shown through air on a diamond (n=2.42) and it partially reflects and refracts.
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You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce
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The time needed for the 7th car to pass is 13.2 s

Explanation:

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And the acceleration is constant so it is

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In attachment the graph of the distance covered versus the time taken: since the motion is uniformly accelerated, the relationship between the two variables is quadratical.

Learn more about accelerated motion:

brainly.com/question/9527152

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brainly.com/question/2562700

#LearnwithBrainly

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